Poj 3624

Charm Bracelet
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 36441 Accepted: 15968
Description

Bessie has gone to the mall’s jewelry store and spies a charm bracelet. Of course, she’d like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a ‘desirability’ factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

Input

• Line 1: Two space-separated integers: N and M
• Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

Output

• Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

Sample Input

4 6
1 4
2 6
3 12
2 7
Sample Output

23

import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int N,M;        int[] w = new int[3410];        int[] p = new int[3410];        int[] dp = new int[13000];//不使用二维数组，超内存        N = in.nextInt();//N件物品        M = in.nextInt();//背包容量        //数据读入        for(int i=1;i<=N;i++){            w[i] = in.nextInt();            p[i] = in.nextInt();        }        for(int i=0;i<N;i++){            dp[i] = 0;        }        for(int i=1;i<=N;i++){            for(int j=M;j>=w[i];j--){                if(dp[j] < dp[j-w[i]]+p[i]){                    dp[j] = dp[j-w[i]]+p[i];                }            }        }        System.out.println(dp[M]);    }}